Calculus Examples

$f (x) = - 3 \log_{4} (x - 3)$

Step 1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 \log_{4} (x - 3)$ with respect to $x$ is $- 3 \frac{d}{d x} [\log_{4} (x - 3)]$ .

$- 3 \frac{d}{d x} [\log_{4} (x - 3)]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \log_{4} (x)$ and $g (x) = x - 3$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $x - 3$ .

$- 3 (\frac{d}{d u} [\log_{4} (u)] \frac{d}{d x} [x - 3])$

Step 2.2

The derivative of $\log_{4} (u)$ with respect to $u$ is $\frac{1}{u \ln (4)}$ .

$- 3 (\frac{1}{u \ln (4)} \frac{d}{d x} [x - 3])$

Step 2.3

Replace all occurrences of $u$ with $x - 3$ .

$- 3 (\frac{1}{(x - 3) \ln (4)} \frac{d}{d x} [x - 3])$

Step 3

Differentiate.

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Step 3.1

Combine $\frac{1}{(x - 3) \ln (4)}$ and $- 3$ .

$\frac{- 3}{(x - 3) \ln (4)} \frac{d}{d x} [x - 3]$

Step 3.2

Move the negative in front of the fraction.

$- \frac{3}{(x - 3) \ln (4)} \frac{d}{d x} [x - 3]$

Step 3.3

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$- \frac{3}{(x - 3) \ln (4)} (\frac{d}{d x} [x] + \frac{d}{d x} [- 3])$

Step 3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- \frac{3}{(x - 3) \ln (4)} (1 + \frac{d}{d x} [- 3])$

Step 3.5

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$- \frac{3}{(x - 3) \ln (4)} (1 + 0)$

Step 3.6

Simplify by multiplying through.

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Step 3.6.1

Add $1$ and $0$ .

$- \frac{3}{(x - 3) \ln (4)} \cdot 1$

Step 3.6.2

Multiply $- 1$ by $1$ .

$- \frac{3}{(x - 3) \ln (4)}$

Step 3.6.3

Apply the distributive property.

$- \frac{3}{x \ln (4) - 3 \ln (4)}$